Ground state of local parent Hamiltonians and invariance under local unitaries Assume that a finite-dimensional pure state $|\psi\rangle\in \mathcal{H}\simeq \mathbb{C}^m$, $m<\infty$, is the (unique) frustration-free ground state of a local parent Hamiltonian and suppose that the locality notion is given in terms of a connected set of neighbourhoods $\{\mathcal{N}_k\}$. My question is the following one: Is it true that any unitary $U$ satisfying $$U|\psi\rangle\langle \psi|U^\dagger=|\psi\rangle\langle \psi|$$ can be decomposed into a finite product of invariance-satisfying unitaries acting only on the neighbourhoods $\{\mathcal{N}_k\}$, that is $U$ can be written as $U=\prod_{i=1}^N U_{\mathcal{N}_{k_i}}$, where every $U_{\mathcal{N}_{k_i}}$ acts only on the neighbourhood $\mathcal{N}_{k_i}$ and it is such that $U_{\mathcal{N}_{k_i}}|\psi\rangle\langle \psi|U_{\mathcal{N}_{k_i}}^\dagger=|\psi\rangle\langle \psi|$
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Any (partial) answer/comment/reference is very welcome.
Thanks in advance.
 A: Isn't simple translation symmetry an example?
E.g. suppose you have a one-dimensional ring of $L$ spins described by $|\psi\rangle = \sum_{\{\sigma_i\}} \psi_{\sigma_1,\sigma_2,\cdots,\sigma_L} \; |\sigma_1,\sigma_2,\cdots,\sigma_N\rangle$. Then this wavefunction could be invariant under the unitary transformation $\psi_{\sigma_1,\sigma_2,\cdots,\sigma_{L-1},\sigma_L} \to \psi_{\sigma_2,\sigma_3,\cdots,\sigma_L,\sigma_1}$. However you clearly cannot do this locally, unless I misunderstand your characterization.
An even more visceral counter-example would be (spatial) inversion symmetry.
A: Consider the toric code Hamiltonian situated on a spherical geometry. This has a unique ground state. Expand the sphere to an infinite radius. Consider a string excitation of the ground state, and loop the string around the sphere (an infinite number of local operations) such that it meets itself, returning our system to its ground state. We know by analogy to the topologically protected ground states of the toric code on an infinite toroidal geometry that such an evolution is not possible with a finite number of local operations. Thus we have described a unitary evolution of the system, with the ground state as its eigenstate, which can not be expressed as a finite decomposition of local operations.
That being said, I may just be cheating by how I am taking the thermodynamic limit, and you may have to clarify such considerations in your question.
A: Here's one idea:
Say $U|\psi\rangle = |\psi\rangle$ for $UU^\dagger = U^\dagger U = I$. 
Then if we consider the exponential form $U = \exp(iG)$ with $G = G^\dagger$ as usual, $|\psi\rangle$ must necessarily be in the kernel  of $G$, $G|\psi\rangle = 0$. 
On the other hand, having $U$ of the product form $U = \prod_k{U_{{\mathcal N}_k} }$ for mutually disjoint neighborhoods is equivalent to $G = \sum_k{G_{{\mathcal N}_k}}$ with $[G_{{\mathcal N}_j}, G_{{\mathcal N}_k}] = 0$ for any ${\mathcal N}_j$, ${\mathcal N}_k$ involved. 
But obviously not all $G$ that have $|\psi\rangle$ in their kernel are of this decomposable form. So not all $U$ such that $U|\psi\rangle = |\psi\rangle$ can be of the product type $U = \prod_k{U_{{\mathcal N}_k} }$.
